Two dimensional complex group representations

Tobias Kildetoft has suggested that I add this as answer, but it still feels too hard:

M.8 Note that since $G$ is simple and non-abelian (else its of prime order) we know that any one-dimensional representation of $G$ is trivial. Now, suppose that $\rho:G\to\text{GL}_2(\mathbb{C})$ is non-trivial. Then, evidently $\rho$ is faithful. But, since it also can't be the sum of one-dimensional reps (since those are all trivial) it must also be irreducible. This implies, in particular, that $2\mid|G|$ so that $G$ has some element of order $2$, say $g$. Note then that $\rho(g)$ must have eigenvalues $\pm 1$ (since it's order $2$). But, as you've already noted, it must also have determinant $1$. This forces the eigenvalues to both be $-1$, and so, in particular, $\rho(g)$ is a scalar matrix, and so in the center of $\text{GL}_2(\mathbb{C})$. By faithfulness, this means that $g$ is a non-trivial element of $Z(G)$, which by simplicity forces $G=Z(G)$. This is a contradiction.

M.14: We may assume that $\rho$ is faithful (just replace $G$ by $G/\ker(\rho)$ and neither the hypothesis nor the conclusion changes). In particular, no non-identity element is in the kernel of $\det$ since $\det(g)$ is precisely the non-one eigenvalue of $g$. Thus $\det:G/\ker(\rho) \to \mathbb{C}^\times$ is an embedding, and $G/\ker(\rho)$ is finite abelian, so every representation is a direct sum of one-dimensional representations.